Tomer is playing a game that involves rolling $5$ $6$ -sided dice at once, and his goal is to roll the dice so exactly $4$ of them land showing a " $1$ ". Which of the following would find the probability that exactly $4$ of the $5$ dice land showing a " $1$ "? Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\dfrac16\right)^4\left(\dfrac56\right)$ (Choice B) B ${6 \choose 5}\left(\dfrac16\right)^4\left(\dfrac56\right)$ (Choice C) C ${6 \choose 5}\left(\dfrac16\right)^4\left(\dfrac56\right)^2$ (Choice D) D ${5 \choose 4}\left(\dfrac16\right)^4\left(\dfrac56\right)$ (Choice E) E ${5 \choose 4}\left(\dfrac16\right)^4\left(\dfrac56\right)^2$
Solution: Probability of $4$ successes We want the probability of getting $4$ successes (die shows " $1$ ") among $5$ dice, so we're going to need $1$ failure (die doesn't not show " $1$ ") as well. The probability of each success is ${\dfrac16}$ and the probability of each failure is $\dfrac56}$. So here's the probability of getting $4$ successes followed by $1$ failure: $\begin{aligned} P(\text{SSSSF})&=\left({\dfrac16}\right)\left({\dfrac16}\right)\left({\dfrac16}\right)\left({\dfrac16}\right)\left(\dfrac56}\right) \\\\ &=\left({\dfrac16}\right)^4\left(\dfrac56}\right) \end{aligned}$ The binomial coefficient ${n \choose k}$ SSSSF isn't the only arrangement that produces $4$ successes among $5$ trials. For instance, SSFSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=4$ successes (die shows " $1$ ") in $n=5$ trials (number of dice), so we should use the binomial coefficient ${5 \choose 4}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $\left(\dfrac16\right)^4\left(\dfrac56\right)$ so for our final answer we multiply this probability by the number of possible arrangements: ${5 \choose 4}\left(\dfrac16\right)^4\left(\dfrac56\right)$ The answer: ${5 \choose 4}\left(\dfrac16\right)^4\left(\dfrac56\right)$